Answer:
x = 3, -3, 2i and -2i where i is an imaginary number.
Step-by-step explanation:
We can express the equation [tex]x^4-5x^2-36=0[/tex] as [tex](x^2)^2-5x^2-36=0[/tex], and we can make [tex]y = x^2[/tex]. So, we have the equation [tex]y^2-5y-36=0[/tex]; to find the solutions of this last equation, we should find two numbers such that these numbers sum up to -5 and multiplied result in -36, these numbers are -9 and 4. Therefore [tex]y^2-5y-36=0[/tex] is equivalent to [tex](y-9)(y+4)=0[/tex], but [tex]y = x^2[/tex], so, [tex](x^2-9)(x^2+4)=0[/tex], i.e., [tex](x+3)(x-3)(x^2+4)=0[/tex]. This last product is equal to zero when x = 3, x = -3 or [tex]x^2=-4[/tex], [tex]x=\pm\sqrt{-4}[/tex], i.e., [tex]x=\pm2i[/tex]. Therefore the solutions of the equation [tex]x^4-5x^2-36=0[/tex] are x = 3, -3, 2i and -2i where i is an imaginary number.
